提取数据问题
[ {
"id": 367,
"playTime": "2022-06-07T03:51:20.000+00:00",
"holeIndex": 1,
"par": 5,
"sort": "1",
"position": "1"
},
{
"id": 367,
"playTime": "2022-06-07T03:51:20.000+00:00",
"holeIndex": 1,
"par": 4,
"sort": "1",
"position": "1"
},
{
"id": 367,
"playTime": "2022-06-07T03:51:20.000+00:00",
"holeIndex": 3,
"par": 3,
"sort": "1",
"position": "1"
},
{
"id": 367,
"playTime": "2022-06-07T03:51:20.000+00:00",
"holeIndex": 4,
"par": 4,
"sort": "1",
"position": "1"
},
{
"id": 367,
"playTime": "2022-06-07T03:51:20.000+00:00",
"holeIndex": 4,
"par": 4,
"sort": "1",
"position": "1"
},
{
"id": 367,
"playTime": "2022-06-07T03:51:20.000+00:00",
"holeIndex": 6,
"par": 5,
"sort": "1",
"position": "1"
},
{
"id": 367,
"playTime": "2022-06-07T03:51:20.000+00:00",
"holeIndex": 7,
"par": 3,
"sort": "1",
"position": "1"
},
{
"id": 367,
"playTime": "2022-06-07T03:51:20.000+00:00",
"holeIndex": 8,
"par": 4,
"sort": "1",
"position": "1"
},
{
"id": 367,
"playTime": "2022-06-07T03:51:20.000+00:00",
"holeIndex": 8,
"par": 4,
"sort": "1",
"position": "1"
}
],
如图，我怎么把相同holeIndex的值取出来放到新的集合中
变成
[ {
"id": 367,
"playTime": "2022-06-07T03:51:20.000+00:00",
"holeIndex": 1,
"par": 5,
"sort": "1",
"position": "1"
},
{
"id": 367,
"playTime": "2022-06-07T03:51:20.000+00:00",
"holeIndex": 1,
"par": 4,
"sort": "1",
"position": "1"
},
],
[
{
"id": 367,
"playTime": "2022-06-07T03:51:20.000+00:00",
"holeIndex": 2,
"par": 5,
"sort": "1",
"position": "1"
},
{
"id": 367,
"playTime": "2022-06-07T03:51:20.000+00:00",
"holeIndex": 2,
"par": 4,
"sort": "1",
"position": "1"
},
]
这种
回答：
将相同属性值取出来放进新的集合，类似这种场景都可以使用stream中的groupingBy来进行分组处理，代码如下
List<Map> mapList = JSON.parseObject(json字符串, List.class); Map<Integer, List<Map>> result = mapList.stream()
.collect(Collectors.groupingBy(map -> (Integer) map.get("holeIndex")));
for (List<Map> list : result.values()) {
System.out.println(JSON.toJSONString(list));
}
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